Is ${827679}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {827679}= &&{8}\cdot100000+ \\&&{2}\cdot10000+ \\&&{7}\cdot1000+ \\&&{6}\cdot100+ \\&&{7}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {827679}= &&{8}(99999+1)+ \\&&{2}(9999+1)+ \\&&{7}(999+1)+ \\&&{6}(99+1)+ \\&&{7}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {827679}= &&\gray{8\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {8}+{2}+{7}+{6}+{7}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${827679}$ is divisible by $3$ if ${ 8}+{2}+{7}+{6}+{7}+{9}$ is divisible by $3$ Add the digits of ${827679}$ $ {8}+{2}+{7}+{6}+{7}+{9} = {39} $ If ${39}$ is divisible by $3$ , then ${827679}$ must also be divisible by $3$ Add the digits of ${39}$ $ {3}+{9} = \color{#9D38BD}{12} $ If $\color{#9D38BD}{12}$ is divisible by $3$ , then ${39}$ must also be divisible by $3$ $\color{#9D38BD}{12}$ is divisible by $3$, therefore ${827679}$ must also be divisible by $3$.